**Q) A train travels a distance of 90 km at a constant speed. Had the speed been 15 km/h more, it would have taken 30 minutes less for the journey. Find the original speed of the train.**

**Ans: **

Let’s consider the speed of the train is X km/hr.

Now, to cover distance of 90 km, it will take hrs

Next, we are given that if speed is increased by 15 km/hr, then new speed will be: (X + 15) km/hr

Now, with this new speed, time taken to cover distance of 90 km, it will take hrs

Given that, new time is 30 mins less i.e. hr less than the present time.

∴

∴

∴ 180 X = (180 – X) ( X + 15)

∴ 180 X = 180 X – X^{2 } + 2700 – 15 X

∴ X^{2 } + 15 X – 2700 = 0

∴ X^{2 } + 60 X – 45 X – 2700 = 0

∴ (X + 60) (X – 45) = 0

∴ X = – 60, X = 45

Since, the speed can not be negative, We reject X = – 60. hence X = 405

**Hence, the speed of the train is 45 km/hr.**

**Check:** At 45 kmph, train will cover 90 km in = 2 hrs.

*By 15 kmph more, new speed is at 60 kmph, and train will take = hrs, *

*Since new time is hr or 30 mins less than earlier – it matches the given condition. Hence, X = 45 kmph is correct.*

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